- 01 May '05 05:56The vertices A and C of a triangle ABC lie on the circumference of a circle. The circle cuts the sides AB and CB at points X and Y respectively. IF AC=XC and Angle YXC = 40 degrees, find the size of Angle ABC.

& (note these are 2 different questions)

In a triangle ABC, Angle ABC = 112 degrees and AB = BC. D is a point on the side AB and E is a point on the side AC such that AE = ED = DB. Find the size of Angle BCD. - 02 May '05 03:06

By my calculation, I get that ang(ABC) should be 45 degrees.*Originally posted by phgao***The vertices A and C of a triangle ABC lie on the circumference of a circle. The circle cuts the sides AB and CB at points X and Y respectively. IF AC=XC and Angle YXC = 40 degrees, find the size of Angle ABC.** - 02 May '05 07:15 / 2 edits

hmmm...actually, i think i made a mistake earlier. I think ABC should be 40 for the first problem.*Originally posted by phgao***Can you provide proof? And what about the 2nd Q?**

My proof of the first question (I think this is right):

We are given that AC = XC. Therefore, it follows (law of sines, for example) that CAX = AXC. (Here my notation is that CAX refers to angle CAX)

So let CAX = AXC = x.

Also, let the radius of the circle be R.

It follows from the law of sines that XC/[sin(x)] = 2R = XC/[sin(XYC)].

It also follows from law of sines that AY/[sin(x+40)] = 2R = AY/[sin(ACY)].

We conclude from these two that:

sin(x) = sin(XYC) and

sin(x+40) = sin(ACY).

This does not mean that x = XYC and x+40 = ACY. In fact, these cannot both hold because then ABC would be 0 which doesn't make sense.

So the only solution I see that is in keeping with YXC = 40 is

XYC = 180-x and ACY = x+40. (for x < 90).

Then to find x, we just set the sum of angles within the circle to 360:

XAC + ACY + XYC + YXA = 360

-> x + (x+40) + (180-x) + (x+40) = 360

-> x = 50

Then it follows that ABC = 180 - XAC - ACY

-> ABC = 180 - (X) - (X+40) = 180 - 50 - 90 = 40

For the second problem, I was getting confused...I kept getting that the problem was not constrained, but I take it that it is, so I need to look at that problem again... - 02 May '05 08:07 / 4 edits

Okay, for the second problem, here is what I get, although I don't have a calculator handy, so I haven't simplified it to the numerical answer:*Originally posted by phgao***In a triangle ABC, Angle ABC = 112 degrees and AB = BC. D is a point on the side AB and E is a point on the side AC such that AE = ED = DB. Find the size of Angle BCD.**

We are given that ABC = 112, and AB = BC. We are also given that AE = ED = DB.

From these, we know that BDC = (68 - BCD)

So we invoke the Law of Sines:

BC/[sin(68 - BCD)] = DB/[sin(BCD)]

Also, AB/[sin(AEB = 129)] = AE/[sin(ABE = 17)].

Then noting that DB = AE and AB = BC, we can simplify these two equations to solve for BCD. When I do that, I get:

cot(BCD) = [(sin(129))/(sin(68)sin(17))] + cot(68)

I don't have a calculator to see what number I get.

Does this look right to you? - 02 May '05 16:33

Now that I have my trusty calculator,*Originally posted by davegage***Okay, for the second problem, here is what I get, although I don't have a calculator handy, so I haven't simplified it to the numerical answer:**

We are given that ABC = 112, and AB = BC. We are also given that AE = ED = DB.

From these, we know that BDC = (68 - BCD)

So we invoke the Law of Sines:

BC/[sin(68 - BCD)] = DB/[sin(BCD)]

Also, AB/ ...[text shortened]... cot(68)

I don't have a calculator to see what number I get.

Does this look right to you?

I get that for Question 2, BCD = 17.